Calorimetry
PHY 213 and PHY 201
Important points to consider:
 Energy entering a body is considered positive and energy
leaving a body is considered negative.
 The temperature of the body must be at its melting temperature
before it can undergo the phase change. The same is true for the
boiling temperature.
 There is no temperature change when a body undergoes a phase
change.
 When two masses are brought into thermal contact, the one at a
higher temperature looses energy (-Q) and that energy flows into
the mass at the lower temperature (+Q).
The Problem:
To 90 g of water contained in a 40-g copper cup, both initially at
40 oC, enough ice, at –15 oC, is added to bring the final
equilibrium temperature to 20 oC. How much ice was added?
This looks like a lesson in grammar. So, I’ll begin by rewording it.
A 40-g copper cup contains 90-g of water. Both cup and water are
at 40 oC. Now, ice, at –15 oC, is added until the temperature of the
entire concoction reaches 20 oC. How much ice (mass) was added?
Ice
Copper cup with water
mice  ?
mwater  0.090 kg
mcopper  0.040 kg
J
c ice  2090
kg oC
J
cwater  4186
kg oC
ccopper
J
 387
o
kg C
L fice
J
 333 k
kg
T0 ice  15 C
o
T0 water  T0copper  40 C
o
T fice  T fcopper  T f water  20 C
o
Of course, this mass will not be
solid at 20 oC.
Assuming the ‘system’ to be isolated from its surroundings, the total
amount of energy in the system is constant. That is, the energy
contained in the ice, water, and copper is constant so we will call it
QT.
Another way of saying this is that the energy going into the ice comes
from the water and copper cup.
Qice  Qwater  Qcopper
Ice gains
energy(+)
Water and copper lose
energy (-)
The water and copper are at a higher temperature than the
ice, so energy flows from them to the ice.
Now, for the ice to get from –15 oC to + 20 oC, there are three
separate processes:
Q
15 0

Qphase

Q020
change
And, to bring the water and copper from 40 oC to 20 oC:
Q  m c T
for both water and copper
 Qice   Qwater  Qcopper

Q

15 0
 Qphase
change

 Q 0 20    Qwater  Qcopper
40 20
40 20
 ice
  mcT copper
 mcT ice  mL f  ice  mcT water   mcT 90g
water
From previous slide:
  mcT copper
 mcT ice  mL f  ice  mcT water   mcT 90g
water
Here, this is the same mass.
Once the ice (the mass we are
solving for) changes phase to
water, we must use “c” for
water. So, now “m” refers to
the mass of the original ice.
This is the original
water in the copper cup.
From previous slide:
  mcT copper
 mcT ice  mL f  ice  mcT water   mcT 90g
water
We will not show all substitutions due to the physical size of
the expression.
 m c (0  15 oC)    m L f    m c (20 oC  0) 
 ice 
 water

 ice 
   m c (20 oC  40)  90 g   m c (20 oC  40) 
copper
water
3
m ice  17 [10 ] kg